Not quite.
Firstly, you assume that there might be finitely many subspaces $w_1,\cdots,w_n$ which contain $S$. This need not always be true. Consider, for instance, the space
$$
P := \{ a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \mid n \in \mathbb{N}, a_i \in \mathbb{R} \}
$$
i.e. the space of all polynomials, interpreted over $\mathbb{R}$ for simplicity. Then the zero vector certainly forms a trivial, singleton subspace, and it is contained in the collections $W_n := \text{span}\{x^n\}$, of which there are infinitely many.
$w_1$,$w_2$...$w_n$ be the subspace spanned by $S$.
Moreover, are you claiming this list is a subspace, or collection of subspaces? It is most definitely not a subspace in itself (since they must be closed under linear combination, and most fields have infinitely many elements).
let W=$w_1$$\cap$$w_2$$\cap$...$\cap$$w_n$. then $W$ is the [...] subspace spanned by $S$ and contain $S$.
If, for some reason, we only have finitely many subspaces $w_1,\cdots,w_n$ that contain $S$, then yes, $W := \bigcap_{i=1}^n w_i$ is the subspace spanned by $S$.
But if you have infinitely many, or in general $\{W_i\}_{i \in I}$ for some arbitrary indexing set $I$, which consists of every subspace containing $S$, then yes:
$$
W := \bigcap_{i \in I} W_i
$$
is the subspace spanned by $S$. More succinctly, if $W \le V$ means $W$ is a vector subspace of $V$,
$$
\text{span}(S) := \bigcap_{\substack{W \le V \\ \text{such that} \\ S \subseteq V}} W
$$
smallest subspace spanned by $S$
There is no "largest" or "smallest" subspace; the subspace spanned by $S$ is unique, there is only one.
You may recall from a more basic linear algebra class that
$$
\text{span}\Big( \{v_1,\cdots,v_n\} \Big) = \left\{ \sum_{i=1}^n c_i v_i \, \middle| \, \text{across all possible choices of } c_i \in \mathbb{R} \right\}
$$
assuming we're working in a real vector space. It may help of you to think this strange definition of the spanning set as this, but generalized in a way that works for infinite sets and infinite-dimensional vector spaces.
